The v^2 versus s graph of a particle moving i | vedclass

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(B) The correct answer is $B$.For uniformly accelerated motion,the equation of motion is $v^2 = u^2 + 2as$,where $v$ is final velocity,$u$ is initial

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Initial velocity of the particle is zero

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(B) The correct answer is $B$.For uniformly accelerated motion,the equation of motion is $v^2 = u^2 + 2as$,where $v$ is final velocity,$u$ is initial velocity,$a$ is acceleration,and $s$ is displacement.Comparing this with the equation of a straight line $y = mx + c$,where $y = v^2$,$x = s$,$m = 2a$ (slope),and $c = u^2$ (intercept on the $v^2$ axis).Since the graph is a straight line,it represents uniformly accelerated motion. This makes statement $A$ correct.Since the graph has a non-zero intercept on the $v^2$ axis,$u^2 
eq 0$,which implies the initial velocity $u 
eq 0$. Therefore,statement $B$ is wrong.For uniformly accelerated motion,$s = ut + \frac{1}{2}at^2$,which is a quadratic equation in $t$,representing a parabola in the $s-t$ graph. Therefore,statement $C$ is correct.Since statement $B$ is wrong,the correct choice is $B$.

The $v^2$ versus $s$ graph of a particle moving in a straight line is shown in the figure. From the graph,some conclusions are drawn. State which statement is wrong?

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